# Open mapping theorem (functional analysis)

Suppose

${displaystyle A:Xto Y}$ is a surjective continuous linear operator. In order to prove that

${displaystyle A}$ is an open map, it is sufficient to show that

${displaystyle A}$ maps the open unit ball in

${displaystyle X}$ to a neighborhood of the origin of

${displaystyle Y.}$ Let

${displaystyle U=B_{1}^{X}(0),V=B_{1}^{Y}(0).}$ Then

${displaystyle X=bigcup _{kin mathbb {N} }kU.}$ Since

${displaystyle A}$ is surjective:

${displaystyle Y=A(X)=Aleft(bigcup _{kin mathbb {N} }kUright)=bigcup _{kin mathbb {N} }A(kU).}$ But

${displaystyle Y}$ is Banach so by Baire’s category theorem

${displaystyle exists kin mathbb {N} :qquad left({overline {A(kU)}}right)^{circ }neq varnothing .}$ That is, we have

${displaystyle cin Y}$ and

${displaystyle r>0}$ ${displaystyle B_{r}(c)subseteq left({overline {A(kU)}}right)^{circ }subseteq {overline {A(kU)}}.}$ Let

${displaystyle vin V,}$ then

${displaystyle c,c+rvin B_{r}(c)subseteq {overline {A(kU)}}.}$ By continuity of addition and linearity, the difference

${displaystyle rv}$ satisfies

${displaystyle rvin {overline {A(kU)}}+{overline {A(kU)}}subseteq {overline {A(kU)+A(kU)}}subseteq {overline {A(2kU)}},}$ and by linearity again,

${displaystyle Vsubseteq {overline {A(LU)}}}$ where we have set

${displaystyle L=2k/r.}$ It follows that for all

${displaystyle yin Y}$ and all

${displaystyle epsilon >0,}$ ${displaystyle xin X}$ such that

${displaystyle qquad |x|_{X}leq L|y|_{Y}quad {text{and}}quad |y-Ax|_{Y}$ Our next goal is to show that

${displaystyle Vsubseteq A(2LU).}$ Let

${displaystyle yin V.}$ By (1), there is some

${displaystyle x_{1}}$ with

${displaystyle left|x_{1}right|$ and

${displaystyle left|y-Ax_{1}right|<1/2.}$ Define a sequence

${displaystyle left(x_{n}right)}$ inductively as follows.
Assume:

${displaystyle |x_{n}|<{frac {L}{2^{n-1}}}quad {text{and}}quad left|y-Aleft(x_{1}+x_{2}+cdots +x_{n}right)right|<{frac {1}{2^{n}}}.qquad (2)}$ Then by (1) we can pick

${displaystyle x_{n+1}}$ so that:

${displaystyle |x_{n+1}|<{frac {L}{2^{n}}}quad {text{and}}quad left|y-Aleft(x_{1}+x_{2}+cdots +x_{n}right)-Aleft(x_{n+1}right)right|<{frac {1}{2^{n+1}}},}$ so (2) is satisfied for

${displaystyle x_{n+1}.}$ Let

${displaystyle s_{n}=x_{1}+x_{2}+cdots +x_{n}.}$ From the first inequality in (2),

${displaystyle left(s_{n}right)}$ is a Cauchy sequence, and since

${displaystyle X}$ is complete,

${displaystyle s_{n}}$ converges to some

${displaystyle xin X.}$ By (2), the sequence

${displaystyle As_{n}}$ tends to

${displaystyle y}$ and so

${displaystyle Ax=y}$ by continuity of

${displaystyle A.}$ Also,

${displaystyle |x|=lim _{nto infty }|s_{n}|leq sum _{n=1}^{infty }|x_{n}|<2L.}$ This shows that

${displaystyle y}$ belongs to

${displaystyle A(2LU),}$ so

${displaystyle Vsubseteq A(2LU)}$ as claimed.
Thus the image

${displaystyle A(U)}$ of the unit ball in

${displaystyle X}$ contains the open ball

${displaystyle V/2L}$ of

${displaystyle Y.}$ Hence,

${displaystyle A(U)}$ is a neighborhood of the origin in

${displaystyle Y,}$ and this concludes the proof.